Queue Using Two Stacks in DSA Rust - Execution Trace

Concept Flow - Queue Using Two Stacks

Enqueue Operation

↓

Push element onto Stack1

↓

Done

↓

Dequeue Operation

↓

Is Stack2 empty?

No→Pop from Stack2

Yes↓

Transfer all elements from Stack1 to Stack2 by popping Stack1 and pushing Stack2

↓

Pop from Stack2

↓

Done

Enqueue adds elements to Stack1. Dequeue pops from Stack2; if Stack2 is empty, elements move from Stack1 to Stack2 first.

Execution Sample

DSA Rust

enqueue(1)
enqueue(2)
dequeue()
enqueue(3)
dequeue()
dequeue()

This code adds 1 and 2, removes one element, adds 3, then removes two elements, showing queue behavior using two stacks.

Execution Table

Step	Operation	Stack1 (top->bottom)	Stack2 (top->bottom)	Action Detail	Queue Visual State
1	enqueue(1)	[1]	[]	Push 1 to Stack1	1 -> null
2	enqueue(2)	[2, 1]	[]	Push 2 to Stack1	1 -> 2 -> null
3	dequeue()	[]	[2, 1]	Stack2 empty, transfer all from Stack1 to Stack2	1 -> 2 -> null
4	dequeue()	[]	[1]	Pop 2 from Stack2	2 -> null
5	enqueue(3)	[3]	[1]	Push 3 to Stack1	2 -> 3 -> null
6	dequeue()	[3]	[]	Pop 1 from Stack2	3 -> null
7	dequeue()	[]	[3]	Stack2 empty, transfer from Stack1 to Stack2	3 -> null
8	dequeue()	[]	[]	Pop 3 from Stack2	null

💡 All operations complete; queue is empty after last dequeue.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8
Stack1	[]	[1]	[2, 1]	[]	[]	[3]	[3]	[]	[]
Stack2	[]	[]	[]	[2, 1]	[1]	[1]	[]	[3]	[]
Queue Size	0	1	2	2	1	2	1	1	0

Key Moments - 3 Insights

Why do we transfer all elements from Stack1 to Stack2 during dequeue when Stack2 is empty?

Because Stack2 holds elements in reverse order for dequeue. Transferring reverses Stack1's order, so popping Stack2 gives the oldest element first (see execution_table rows 3 and 4).

Why don't we push new elements directly to Stack2?

New elements go to Stack1 to keep enqueue fast. Stack2 is only for dequeue order. This separation keeps operations efficient (see execution_table rows 1, 2, and 5).

What happens if we dequeue when both stacks are empty?

The queue is empty, so dequeue cannot remove elements. This case is not shown here but would return an error or None.

Visual Quiz - 3 Questions

Look at the execution_table at Step 4, what is the top element of Stack2 after popping?

A 3

B 2

C 1

D Empty

💡 Hint

Check Stack2 column at Step 4 after pop operation.

At which step does Stack2 become empty after a dequeue operation?

A Step 3

B Step 6

C Step 7

D Step 8

💡 Hint

Look at Stack2 column and dequeue operations in execution_table.

If we enqueue(4) after Step 6, what will be the state of Stack1?

A [4]

B [3,4]

C [4,3]

D []

💡 Hint

Stack1 after Step 6 is [3], enqueue pushes new element on top.

Concept Snapshot

Queue Using Two Stacks:
- Enqueue: push element onto Stack1
- Dequeue: if Stack2 empty, move all from Stack1 to Stack2
- Then pop from Stack2
- Stack1 stores newest elements, Stack2 stores oldest
- Efficient queue with two stacks

Full Transcript

This visualization shows how a queue can be implemented using two stacks. Enqueue operations push elements onto Stack1. Dequeue operations pop elements from Stack2. If Stack2 is empty during dequeue, all elements from Stack1 are moved to Stack2 to reverse their order, ensuring the oldest element is dequeued first. The execution table tracks each step's operation, the state of both stacks, and the visual queue order. Variable tracking shows how Stack1, Stack2, and queue size change after each step. Key moments clarify why elements move between stacks and how the queue maintains order. The quiz tests understanding of stack states and operations during the process.

Practice

1. What is the main idea behind implementing a Queue Using Two Stacks?

easy

A. Use one stack for enqueue and another for dequeue operations

B. Use two stacks to perform only enqueue operations

C. Use two stacks to store elements in sorted order

D. Use two stacks to reverse the order twice to maintain queue order

Solution

1. Step 1: Understand queue order

   A queue follows First-In-First-Out (FIFO) order, meaning the first element added is the first removed.

2. Step 2: Use two stacks to reverse order twice

   One stack reverses the order when pushing elements, and the second stack reverses it back when popping, preserving FIFO.

3. Final Answer:

   Use two stacks to reverse the order twice to maintain queue order -> Option D

4. Quick Check:

   Two stacks reverse order twice = queue order [OK]

Hint: Two stacks reverse order twice to simulate queue [OK]

Common Mistakes:

• Thinking both stacks store elements in sorted order
• Using stacks only for enqueue or dequeue
• Confusing stack order with queue order

2. Which of the following Rust code snippets correctly declares two stacks for implementing a queue?

easy

A. let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new();

B. let stack1 = Vec<i32>; let stack2 = Vec<i32>;

C. let mut stack1 = Vec::new<i32>(); let mut stack2 = Vec::new<i32>();

D. let stack1: Vec = Vec::new(); let stack2: Vec = Vec::new();

Solution

1. Step 1: Check correct Vec declaration syntax in Rust

   In Rust, to declare a mutable vector of i32, use: let mut stack: Vec<i32> = Vec::new();

2. Step 2: Compare options with correct syntax

   let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new(); matches the correct syntax exactly; others have syntax errors or missing mutability.

3. Final Answer:

   let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new(); -> Option A

4. Quick Check:

   Correct Rust Vec declaration = let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new(); [OK]

Hint: Use let mut stack: Vec<i32> = Vec::new() for mutable stacks [OK]

Common Mistakes:

• Omitting mut keyword for mutable vectors
• Incorrect generic syntax like Vec<i32> without parentheses
• Using Vec::new<i32>() which is invalid

3. What will be the output of the following Rust code implementing a queue using two stacks?

struct Queue {
    stack1: Vec<i32>,
    stack2: Vec<i32>,
}

impl Queue {
    fn new() -> Self {
        Queue { stack1: Vec::new(), stack2: Vec::new() }
    }

    fn enqueue(&mut self, val: i32) {
        self.stack1.push(val);
    }

    fn dequeue(&mut self) -> Option<i32> {
        if self.stack2.is_empty() {
            while let Some(x) = self.stack1.pop() {
                self.stack2.push(x);
            }
        }
        self.stack2.pop()
    }
}

fn main() {
    let mut q = Queue::new();
    q.enqueue(10);
    q.enqueue(20);
    q.enqueue(30);
    println!("{:?}", q.dequeue());
    println!("{:?}", q.dequeue());
    q.enqueue(40);
    println!("{:?}", q.dequeue());
    println!("{:?}", q.dequeue());
}

medium

A. Some(10) Some(20) Some(30) Some(40)

B. Some(30) Some(20) Some(10) Some(40)

C. Some(10) Some(20) Some(40) None

D. None None None None

Solution

1. Step 1: Trace enqueue operations

   Enqueue 10, 20, 30 pushes these onto stack1 in order: [10, 20, 30]

2. Step 2: Trace dequeue operations

   First dequeue: stack2 empty, pop all from stack1 to stack2: stack2 = [30, 20, 10]. Pop from stack2 returns 10.
   Second dequeue: pop from stack2 returns 20.
   Enqueue 40: push 40 to stack1.
   Third dequeue: pop from stack2 returns 30.
   Fourth dequeue: stack2 empty, move 40 from stack1 to stack2, pop returns 40.

3. Final Answer:

   Some(10) Some(20) Some(30) Some(40) -> Option A

4. Quick Check:

   Queue order preserved by two stacks = Some(10) Some(20) Some(30) Some(40) [OK]

Hint: Dequeue empties stack2 by reversing stack1 first [OK]

Common Mistakes:

• Assuming stack2 pops in same order as stack1
• Not moving elements from stack1 to stack2 when stack2 is empty
• Confusing order of elements after transfer

4. Identify the bug in this Rust code snippet for dequeue operation in a queue using two stacks:

fn dequeue(&mut self) -> Option<i32> {
    while let Some(x) = self.stack1.pop() {
        self.stack2.push(x);
    }
    self.stack2.pop()
}

medium

A. It uses pop() on stack1 instead of peek()

B. It moves elements from stack1 to stack2 every time, even if stack2 is not empty

C. It does not pop from stack2 after moving elements

D. It returns None always because stack2 is never filled

Solution

1. Step 1: Analyze the dequeue logic

   The code moves all elements from stack1 to stack2 every dequeue call, regardless of stack2's state.

2. Step 2: Identify correct behavior

   To maintain queue order efficiently, elements should be moved only if stack2 is empty. Otherwise, stack2's elements should be popped directly.

3. Final Answer:

   It moves elements from stack1 to stack2 every time, even if stack2 is not empty -> Option B

4. Quick Check:

   Move only if stack2 empty = It moves elements from stack1 to stack2 every time, even if stack2 is not empty [OK]

Hint: Move elements only if stack2 is empty during dequeue [OK]

Common Mistakes:

• Moving elements every dequeue wastes time
• Ignoring stack2's current elements
• Not popping from stack2 after transfer

5. You want to implement a peek function for the queue using two stacks in Rust that returns the front element without removing it. Which approach correctly achieves this?

hard

A. Return the first element of stack1 without any stack transfers

B. Return the last element of stack1 directly without moving elements

C. If stack2 is empty, move all elements from stack1 to stack2, then return the last element of stack2 without popping

D. Pop an element from stack2 and push it back immediately to peek

Solution

1. Step 1: Understand peek operation in queue using two stacks

   Peek should return the front element without removing it. The front is at the bottom of stack1 or top of stack2 after transfer.

2. Step 2: Correct approach for peek

   If stack2 is empty, move all elements from stack1 to stack2 to reverse order. Then peek the last element of stack2 (top) without popping.

3. Final Answer:

   If stack2 is empty, move all elements from stack1 to stack2, then return the last element of stack2 without popping -> Option C

4. Quick Check:

   Peek uses stack2 top after transfer if empty = If stack2 is empty, move all elements from stack1 to stack2, then return the last element of stack2 without popping [OK]

Hint: Peek top of stack2 after transfer if empty [OK]

Common Mistakes:

• Peeking stack1 directly ignores queue order
• Popping to peek then pushing back wastes time
• Ignoring stack2 state before peeking